Is it hard to tackle the integral ∫ 0 ∞ x 2 ( 1 + x 4 ) 2 d x ?

Question

Is it hard to tackle the integral       ∫          0              ∞                  x              2                            (        1        +                  x                      4                          )                    2              d  x  ?

x713x9o7r · Accepted Answer

Integrate both sides of the equation             (                        x          3                          1          +                      x            4                              )        ′    =            4              x        2                    (      1      +              x        4                    )        2              −            x      2              1      +              x        4             to obtain      ∫          0              ∞                  x              2                    (      1      +              x                  4                            )                  2                      d  x  =      1    4        ∫          0              ∞                  x              2                    1      +              x                  4                      d  x  =      1    4    ⋅      π          2              2              =      π          8              2

trumansoftjf0 · Accepted Answer

You can use residue theorem. Start by extending the integral from

- \infty

to

+ \infty

, which you can do since the function is even. Then, use the semi-circle in the complex plane for integration:

Γ (θ) = R e^{i θ}

, with

θ \in [0, π]

. You will get the desired result in the limit

R \to + \infty

. In that limit, the integral in the semi-circle vanishes due to Jordan Lemma. The residues inside the contour are

e^{i π / 4}

and

e^{i 3 π / 4}

, with Res

(\frac{z^{2}}{(z^{4} + 1)^{2}}, z = e^{i π / 4}) = \frac{e^{3 i π / 4}}{16}

and Res

(\frac{z^{2}}{(z^{4} + 1)^{2}}, z = e^{i 3 π / 4}) = \frac{e^{5 i π / 4}}{16}

. The result is then

\int_{0}^{+ \infty} d x \frac{x^{2}}{(x^{4} + 1)^{2}} = \frac{1}{2} \int_{- \infty}^{+ \infty} d x \frac{x^{2}}{(x^{4} + 1)^{2}} = π i \sum Res = \frac{π i}{8} (e^{5 i π / 4} + e^{3 i π / 4}) = \frac{π}{8 \sqrt{2}} = \frac{\sqrt{2} π}{16}

Is it hard to tackle the integral int_0^(oo) (x^2)/((1+x^4)^2)dx?

Answered question

Answer & Explanation

New Questions in Trigonometry